Intermolecular Forces

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1 Intermolecular Forces: Introduction Intermolecular Forces Forces between separate molecules and dissolved ions (not bonds) Van der Waals Forces 15% as strong as covalent or ionic bonds Chapter 11 Intermolecular Forces: Introduction Low temperature strong High temperature kinetic energy of motion overcomes the IMF Boiling point is a good indicator Stronger IMF = higher boiling point Weaker IMF = lower boiling point Ion-Ion Ion-Dipole Full to Full Charge Full to Partial Charge Hydrogen Bonds Partial to Partial Charge (H involved) Dipole-Dipole Partial to Partial Charge (No H) London Dispersion Forces Non-polar to Non-Polar Predict what type of IMF would form between: a. Br 2 and I 2 b. KCl and water c. Water and ammonia d. Two SO 2 molecules e. NaCl ions in a crystal Type 1: Ion-Ion Forces Full Charges to full charges High melting points (ionic solids) Ex: Na NaCl Melting Point ~98 o C ~800 o C 1

2 Type 2: Ion-Dipole Forces Full Charges to partial charges Very Strong Type 3: Hydrogen Bonds Stronger than dipole-dipole that do not have hydrogen (no inner electrons, strong + pull) Generally involves hydrogen and O, N or F R-H O-R R-H N-R R-H F-R Miscibility Miscibility Like dissolves like Substances that can hydrogen bond dissolve in one another. 2

3 Glucose and other sugars Water Beading Ice Ice DNA 3

4 DNA is TWO molecules that are hydrogen bonded (like a zipper) DNA Generally increases with increasing molar mass H 2 O unusually high - H-bonding Boiling Point Type 4: Dipole-Dipole Slightly weaker IMF Involve + and - charges other than those in hydrogen bonding Draw Lewis Dot Structures to explain the following boiling points MM (amu) BP ( o C) CH 3 CH 2 CH CH 3 CHO CH 3 CN Type 5: London Dispersion Forces Very weak IMF Caused by temporary imbalances in electrons London Forces: Inorganic Molecules More electrons, more chance for temporary dipole Boiling Point Table Halogen Molar Mass BP( o C) Noble Gas Atomic Mass BP( o C) F 2 (g) He Cl 2 (g) Ne Br 2 (l) Ar I 2 (s) Kr

5 Explain the differences in boiling point between Cl 2 (-35 o C) and Ar (-186 o C) London Forces: Organic Compounds The longer the carbon chain, the higher the London Dispersion Forces (the higher the melting point and boiling point) Chainlike molecules greater London Forces than bunched up molecules (branched) Ex 1 Rank the following compounds in terms of increasing melting points: NaCl, CF 4, CH 3 OH. Ex 2 Separate the following compounds by whether they have dipole-dipole attractions (including H- bonding) or London Forces. Which should have the highest dipole-dipole attraction? Which should have the strongest London Force? Ex 3 Rank the following in order of increasing boiling point: BaCl 2, H 2, CO, HF, Kr Br 2, Ne, HCl, N 2, HF 5

6 Ex 4 Rank the following in order of increasing boiling point: N 2, KBr, O 2, CH 3 CH 2 OH, HCN Properties of Liquid Viscosity resistance of a liquid to flow Oil is more viscous than water Water has H-bonds (stronger) Oil has London forces (weaker, but there are many more of them, long carbon chain) Miscibility 40 W oil 10 W oil Surface Tension Surface Tension 6

7 Capillary Action Water is attracted to the glass Mercury more attracted to itself Heating Curves 1. Changes of state do not have a temperature change. 1. Melting/Freezing 2. Boiling/Condensing 2. A glass of soda with ice will stay at 0 o C until all of the ice melts. 3. Graph flattens out during changes of state Heating Curves Heating Curve Temperature ( o C) Melting Ice warms up Boiling Water warms up Steam heats up No phase change is occurring (heating ice, water, or steam): q = mc p Melting or boiling: T q = ml f or q = ml v Heat (Joules) L f and L v Heating Curves Latent Heat - heat for phase changes. No temperature change. L f latent heat of freezing/melting Temperature ( o C) Use q = ml v Boiling L v latent heat of boiling/condensing Use q = ml f Melting Use q = mc p T Heat (calories) 7

8 Substance C p Steam 2.01 J/g o C Water 4.18 J/g o C Ice 2.09 J/g o C Important Values Heating Curves: Example 1 How much energy must be removed to cool grams of water at 20.0 o C to make ice at 10.0 o C? Latent Heat of fusion (water) L f = J/g Latent Heat of vaporization(water) L v = J/g Heating Curves: Example 1 Heating Curves: Example 1 Temperature ( o C) Melting (q=ml f ) Water cools (q=mc p T) Ice cools (q=mc p T) Heat 1. Cooling the water q = mc p T = (100 g)(4.18 J/g o C)(0 o C-20 o C) q = 8360 J (8.36 kj) (ignore the negative sign for now) 2. Freezing the water q = ml f = (100.0 g)(334.7 J/g)= kj 3. Cooling the ice down to 10.0 o C q = mc p T = (100 g)(2.09 J/g o C)(-10 o C-0 o C) q = 2.09 kj (we will ignore the negative sign for now) 8.36 kj kj kj= kj Ex 2 Ex 3 How much energy is needed to convert 18.0 grams of ice at -25 o C to steam at 125 o C? How much energy must be used to convert grams of water from steam at o C to ice at o C? ANS: 56.0 kj (309 kj) 8

9 Vapor Pressure Pressure of a gas above a liquid caused by that liquid Temperature measure of the average kinetic energy of molecules At any given moment, some molecules have enough energy to escape EX: Even cold water will evaporate Volatility Volatile liquids that evaporate easily Acetone Often weak intermolecular forces Boiling point point at which the vapor pressure of a liquid = vapor pressure of the atmosphere Normal Boiling Point vapor pressure = 1 atm Steam pressure cookers Forces water to boil at a higher temperature High Altitude water boils at a lower temperature Four Types of Solids 1. Molecular Solids (single molecules) 2. Covalent Network Solids (one large molecule) 3. Ionic Solids 4. Metallic Solids Held together by IMF Ice Plastics 1. Molecular Solids 9

10 2. Covalent Network Solids Basically one big molecule Held together by covalent bonds Diamond Graphite Quartz (SiO 2 ) 3. Ionic Solids Held together by electrostatic attraction (Ion:Ion) Usually crystalline (unit cells) 4. Metallic Solids Atoms share electrons very freely Positive nuclei in a sea of electrons Electrons held loosely Conducts electricity Photoelectric effect Malleable and ductile Rank by boiling point (low to high). Below each, tell me which IMF is important: CH 3 OH Cl 2 N 2 CH 3 Cl CH 3 CH 2 CH 2 CH 3 CH 3 CH 2 CH 2 OH Draw Lewis Dots for these CH 3 OCH 3 CH 3 CH 2 CH 3 10

11 (100.0g)(2.01 J/gK)(20.0 o C) = 4.02 kj (100.0g)( J/g) = kj (100.0g)(4.18 J/gK)(100.0 o C) = 41.8 kj (100.0g)(334.7 J/g) = kj (100.0g)(2.09 J/gK)(5.0 o C) = 1.05 kj kj Water q cool (100)(4.18)(10) = 4.18 kj q freeze (100)(334.7) = kj kj Ammonia q v = ml v m =q v /L v m = kj/23.35 kj/mol = 1.61 mol mass = (1.61 mol)(17.0 g/mol) = 27.4 g 2. a) H-bonding, (b)london (c) Ion-dipole (d) dipole-dipole. Ion-dipole and h-bonding are stronger 10. a) Solids = attractive forces (IMF) win Liquids = Balance Gases = Kinetic energy wins b) Increasing T increase KE, eventually overcoming IMF c) High pressure forces gas molecules clsoe together and IMF s can win 12.a) Distance greater in liquid state b) More movement, more volume, lower density 14. Overall, net forces are attractive 16.a) CH 3 OH has h-bonding, CH 3 SH does not b) Xe is heavier, greater London Forces c) Cl 2 more polarizable than Kr d) Acetone has dipole-dipole forces 18. a) True b) False c) False d) True 20. a) Br 2 b) C 5 H 11 SH c) CH 3 CH 2 CH 2 Cl 22. Propyl alcohol is longer and more polarizable 24. a) HF has hydrogen bonds, HCl dipole/dipole b) CHBr 3 higher molar mass, more dispersion c) ICl has dipole-dipole, Br 2 only dispersion 26.a) Dispersion, C 8 H 18 higher boiling point b) C 3 H 8 (dispersion) CH 3 OCH 3 (dip-dip) c) HOOH (h-bonding) HSSH (dip-dip) d) NH 2 NH 2 (h-bonding) CH 3 CH 3 (dispersion) 32. H 2 NNH 2, HOOH, H 2 O can all h-bond 11

12 34.a) Exo b) Endo c) Endo d) Exo g CCl 2 F kj 12

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